The number of solutions of the equation `e^(sin x) - 2e^(-sin x) = 2` is

Options

A. 1
B. 0
C. infinite
D. 2

Answer

To solve the equation \( e^{\sin x} - 2e^{-\sin x} = 2 \), we will follow these steps:

### Step 1: Substitute \( e^{\sin x} \)
Let \( t = e^{\sin x} \). Then, the equation becomes:
\[ t - 2 \cdot \frac{1}{t} = 2 \]

### Step 2: Multiply through by \( t \)
To eliminate the fraction, multiply both sides by \( t \) (noting that \( t > 0 \)):
\[ t^2 - 2 = 2t \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ t^2 - 2t - 2 = 0 \]

### Step 4: Use the quadratic formula
Now we can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -2, c = -2 \):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 + 8}}{2} \]
\[ t = \frac{2 \pm \sqrt{12}}{2} \]
\[ t = \frac{2 \pm 2\sqrt{3}}{2} \]
\[ t = 1 \pm \sqrt{3} \]

### Step 5: Evaluate the solutions for \( t \)
The two possible values for \( t \) are:
\[ t_1 = 1 + \sqrt{3} \quad \text{and} \quad t_2 = 1 - \sqrt{3} \]

### Step 6: Determine the validity of the solutions
Since \( t = e^{\sin x} \) must be positive, we need to check the values:
- \( t_1 = 1 + \sqrt{3} \) is positive.
- \( t_2 = 1 - \sqrt{3} \) is negative (since \( \sqrt{3} \approx 1.732 \), thus \( 1 - \sqrt{3} < 0 \)).

### Step 7: Find \( \sin x \)
Now we only consider \( t_1 \):
\[ e^{\sin x} = 1 + \sqrt{3} \]
Taking the natural logarithm:
\[ \sin x = \ln(1 + \sqrt{3}) \]

### Step 8: Check the range of \( \sin x \)
The value of \( \ln(1 + \sqrt{3}) \) needs to be checked against the range of \( \sin x \), which is \([-1, 1]\):
- Calculate \( \ln(1 + \sqrt{3}) \):
- \( 1 + \sqrt{3} \approx 2.732 \)
- \( \ln(2.732) \approx 1.0 \)

Since \( \ln(1 + \sqrt{3}) \) is approximately \( 1.0 \), it is at the upper limit of the range of \( \sin x \).

### Conclusion
Since \( \sin x = \ln(1 + \sqrt{3}) \) is equal to \( 1 \), it corresponds to a unique solution \( x = \frac{\pi}{2} + 2k\pi \) for any integer \( k \). However, since \( \sin x \) cannot exceed \( 1 \), we conclude that there are no valid solutions.

Thus, the number of solutions to the equation \( e^{\sin x} - 2e^{-\sin x} = 2 \) is:
\[ \boxed{0} \]